bitset

关于 set ，必须说明的是 set 关联式容器。 set 作为一个容器也是用来存储同一数据类型的数据类型，并且能从一个数据集合中取出数据，在 set 中每个元素的值都唯一，而且系统能根据元素的值自动进行排序。应该注意的是 set 中数元素的值不能直接被改变。 C++ STL 中标准关联容器 set, multiset, map, multimap 内部采用的就是一种非常高效的平衡检索二叉树：红黑树，也成为 RB 树 (Red-Black Tree) 。 RB 树的统计性能要好于一般平衡二叉树，所以被 STL 选择作为了关联容器的内部结构。 注意： 1 、 set 中的元素都是排好序的 2 、 set 集合中没有重复的元素 来自 <https: //blog.csdn.net/byn12345/article/details/79523516> bitset C++ 的 bitset 在 bitset 头文件中，它是一种类似数组的结构，它的每一个元素只能是０或１，每个元素仅用１ bit 空间。 bitset<4> bitset1; 　　 // 无参构造，长度为４，默认每一位为０ bitset<8> bitset2(12); 　　 // 长度为８，二进制保存，前面用０补充 string s = "100101" ; bitset<10> bitset3(s); 　　 // 长度为

How do I read a binary file into a bitset or vector<bool> ? The binary file will vary in length. Is there a better container for this? I am new to C++ though experienced as a programmer. If the file is large, Why should you read once, whole the file into the memory? You can read a little piece every time. The size is determined with the size in this func: file.read(buff, size) When the buff is char's array. I'm sorry, but You can't simplest read/save vector to file. for more details see here and here . And use Google, It's very helpful... You didn't give too much context of what you're trying

Am I missing something painfully obvious? Or does just nobody in the world actually use java.util.BitSet? The following test fails: @Test public void testBitSet() throws Exception { BitSet b = new BitSet(); b.set(0, true); b.set(1, false); assertEquals(2, b.length()); } It's really unclear to me why I don't end up with a BitSet of length 2 and the value 10. I peeked at the source for java.util.BitSet, and on casual inspection it seems to fail to make sufficient distinction between a bit that's been set false and a bit that has never been set to any value... (Note that explicitly setting the

I have a need for a BitSet which allows easy concatenation of multiple BitSets create a new BitSet. The default implementation doesn't have such a method. Is there any an implementation in some external library that any of you know which allows easy concatenation? For example lets say I have a bitarray 11111 and another bit array 010101. I want functionality like appending. So after concatenating it would result in 11111010101. Well there's no way to implement this terribly efficient (performance and memory that is) since there's no leftshift method. What you can do is either use the obvious

I am looking for an efficient way how to easily convert a BitSet to a binary string. Let us say that its usual length would be thousands of bits. For example, lets have this: BitSet bits = new BitSet(8); bits.set(1); bits.set(3); And this is the desired result: String result = toBinaryString(bits); // expected: result = "01010000" I have some ideas in general (streams, etc.), but there might be some obvious standard method I am just missing. So this is the most efficient way I have tried so far: private static class FixedSizeBitSet extends BitSet { private final int nbits; public

I have a problem that I am a bit stuck on and I was informed by a colleague that this would be a good place to seek help. I am trying to implement a C style bitfield in Java. Here is a rough example (I do not have the actual code in front of me at this moment). typedef union { typedef struct { unsigned short a :1; unsigned short b :1; unsigned short c :2; unsigned short d :10; } bitfield; unsigned short bitmap; }example_bitfield; I have a good bit of similar style bitfields from legacy code. The reason that I need to come up with an equivalent method for Java is that I am working on code that

Any quick method to count the number of set bits in a BitSet other than the usual 'keep a counter' method? pwc The cardinality() method returns the number of set bits. (Assuming you don't want to call cardinality()) int count = 0; for (int i = bs.nextSetBit(0); i >= 0; i = bs.nextSetBit(i+1)) { count++; } see javadoc Aditya Parmar BitSet B1 = new BitSet(3); B1.set(0); B1.cardinality(); Output: 1 来源： https://stackoverflow.com/questions/4883172/java-count-number-of-bits-set-in-a-java-util-bitset